The time needed for Q to reach 20% of its final value

[Fatima Hasan]

Homework Statement

Homework Equations

Q = Q_max (1-e^-t/τ)
Q= CΔV

The Attempt at a Solution

CΔV = CΔV (1-e^-t/RC)
0.2 = (1-e^{-t/(1*106*10*10-6})
ln (0.2) = $-\frac{t}{10}$
t = - ln(0.2)*10
= 16.09 s
Can someone tell me where is my mistake ?

 

[magoo]

What is the final value of charge on the capacitor?

What is 20% of that final value?

 

[Fatima Hasan]

What is 20% of that final value?

20% Q_f = 20%* C*ΔV
= 20% * 10*10^-6 *10
= 2*10^-5 C

 

[haruspex]

What is the final value of charge on the capacitor?

No need to find that. Fatima's method was essentially correct.

 

[haruspex]

CΔV = CΔV (1-e^-t/RC)
0.2 = (1-e^{-t/(1*106*10*10-6})

How are you calculating R?

 

[haruspex]

There are two resistors in the circuit , but they are not connected in series nor in parallel . I calculated R as it given in the question = 1 MΩ.

well, they are both in the circuit. When a current I flows, what is the consequence of them for the potential across the capacitor?

 

[magoo]

In addition to the resistance value that haruspex pointed out, your equation

0.2 = 1 - e^(-t/RC)

needs to be rearranged before you take the natural logs.

 

[Fatima Hasan]

In addition to the resistance value that haruspex pointed out, your equation

0.2 = 1 - e^(-t/RC)

needs to be rearranged before you take the natural logs.

0.2 = 1 - e^{(t)/(2*106*10*10-6)}
e^-t/τ = 0.8
ln (0.8) = - t / 20
t = - ln (0.8) *20
= 4.46 s

 

[magoo]

Nice work!